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POJ 1995 Raising Modulo Numbers(快速幂)

作者:互联网

嗯...

 

题目链接:http://poj.org/problem?id=1995

 

快速幂模板...

 

AC代码:

 1 #include<cstdio>
 2 #include<iostream>
 3 
 4 using namespace std;
 5 
 6 int main(){
 7     long long N, M, n, a, b, c, sum = 0;
 8     scanf("%lld", &N);
 9     while(N--){
10         scanf("%lld%lld", &M, &n);
11         sum = 0;
12         for(int i = 1; i <= n; i++){
13             c = 1;
14             scanf("%lld%lld", &a, &b);
15             while(b){
16                 if(b & 1) c = c * a % M;
17                 a = a * a % M;
18                 b /= 2;
19             }
20             sum += c % M;
21         }
22         printf("%lld\n", sum % M);
23     }
24     return 0;
25 }
AC代码

 

标签:...,1995,int,sum,long,POJ,Numbers,include,lld
来源: https://www.cnblogs.com/New-ljx/p/11515343.html